How do you find the limit of $sin (\frac{x - 1}{2 + x^{2}})$ $sin((x-1)/(2+x^2))$ as x approaches infinity?

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Answer 1 · 2016-12-08T03:37:45

$lim_(x->oo)sin((x-1)/(2+x^2))=0$

Because $f(x)=sin(x)$ is continuous, we have

$lim_(x->oo)sin((x-1)/(2+x^2)) = sin(lim_(x->oo)(x-1)/(2+x^2))$

$=sin(lim_(x->oo)(1/x-1/x^2)/(2/x^2+1))$

$=sin((0-0)/(0+1))$

$=sin(0)$

$=0$

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